The Arc Length of a Smooth, Planar Curve and Distance Traveled

AP Calculus BC · Applications of Integration

Arc Length and Distance Traveled

Pythagoras on an infinitesimal step, integrated from one end of the curve to the other.

key idea$L=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\;dx$

Picture the region

Arc length is the Pythagorean theorem applied to an infinitesimal step. Move $dx$ across and $dy$ up, and the piece of curve you traverse has length $ds=\sqrt{dx^{2}+dy^{2}}$. Factoring $dx$ out of that square root produces the formula.

One magnified step along $y=\dfrac{x^{2}}{3}$. The horizontal leg is $dx$, the vertical leg is $dy$, and the hypotenuse $ds$ is the sliver of curve. Since $\dfrac{dy}{dx}$ is the slope, $ds=\sqrt{1+\left(\dfrac{dy}{dx}\right)^{2}}\,dx$ — and the integral adds up every such sliver.

[Figure: Arc length is the limit of a chain of straight chords. Each contributes root(dx2+dy2), and letting the chords shrink turns that sum into the integral.]

Arc length is the limit of a chain of straight chords. Each contributes $\sqrt{dx^2+dy^2}$, and letting the chords shrink turns that sum into the integral.

Setting up the integral

Building an arc-length integral

Confirm the curve is smooth on the interval — $f'$ must exist and be continuous. A corner or a vertical tangent breaks the formula.
Differentiate to get $\dfrac{dy}{dx}$, then square it. Almost all the work of these problems lives in this step.
Add $1$ and take the square root. Simplify only if a perfect square appears — most integrands do not simplify at all.
Set the limits as the $x$-values of the endpoints, and integrate.
If the result is not elementary, say so and evaluate numerically. On the exam these appear on the calculator section for exactly that reason.

Smooth curveOne whose derivative exists and is continuous on the interval — no corners, cusps, or breaks.

$ds$The infinitesimal arc element, $\sqrt{dx^{2}+dy^{2}}$. Every arc-length formula is a way of writing $\displaystyle\int ds$.

Distance travelledFor motion along a curve, the arc length of the path — unsigned, so reversals still add.

Perfect-square integrandThe rare case where $1+\left(f'\right)^{2}$ factors, making the integral elementary. Exam problems are built around these.

Curve given as	Formula	Integrate over
$y=f(x)$	$\displaystyle\int\sqrt{1+\left(f'(x)\right)^{2}}\,dx$	$x$ from $a$ to $b$
$x=g(y)$	$\displaystyle\int\sqrt{1+\left(g'(y)\right)^{2}}\,dy$	$y$ from $c$ to $d$
a straight segment	$\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}$	no integral needed
a curve with a vertical tangent	—	switch to the $dy$ form

Rows 1 and 2 are the same formula seen from two directions. Choose whichever variable makes the derivative simpler — and row 4 is a genuine requirement, since a vertical tangent makes $f'$ undefined but $g'$ perfectly well behaved.

Worked examples

EX 1

A perfect-square integrand

$y=\tfrac23x^{3/2}$

Find the length of $y=\dfrac23x^{3/2}$ from $x=0$ to $x=3$.

Differentiate

$\dfrac{dy}{dx}=x^{1/2}$.

Square and add one

$1+\left(x^{1/2}\right)^{2}=1+x$.

Set up

$L=\displaystyle\int_{0}^{3}\sqrt{1+x}\,dx$.

Integrate

$\left[\dfrac23(1+x)^{3/2}\right]_{0}^{3}=\dfrac23\left(8-1\right)$.

Evaluate

$L=\dfrac{14}{3}$. The exponent $\tfrac32$ was chosen precisely so the square root would disappear.

EX 2

Verify against a known length

a straight line

Find the length of $y=3x+1$ from $x=0$ to $x=4$, and check it geometrically.

Differentiate

$\dfrac{dy}{dx}=3$, a constant.

Build the integrand

$\sqrt{1+9}=\sqrt{10}$, also constant.

Integrate

$L=\displaystyle\int_{0}^{4}\sqrt{10}\,dx=4\sqrt{10}$.

Check with the distance formula

$\Delta x=4$ and $\Delta y=12$, so the length is $\sqrt{16+144}=\sqrt{160}=4\sqrt{10}$. ✓

Note the reassurance

The formula reduces to the Pythagorean distance for a straight line, which is exactly what it was built from.

EX 3

Integrating in $y$ instead

avoid the vertical tangent

Find the length of $x=\dfrac{y^{3}}{6}+\dfrac{1}{2y}$ from $y=1$ to $y=2$.

Differentiate with respect to $y$

$\dfrac{dx}{dy}=\dfrac{y^{2}}{2}-\dfrac{1}{2y^{2}}$.

Square it

$\dfrac{y^{4}}{4}-\dfrac12+\dfrac{1}{4y^{4}}$.

Add one

$\dfrac{y^{4}}{4}+\dfrac12+\dfrac{1}{4y^{4}}=\left(\dfrac{y^{2}}{2}+\dfrac{1}{2y^{2}}\right)^{2}$ — a perfect square, because adding $1$ flipped the middle sign.

Take the root and integrate

$L=\displaystyle\int_{1}^{2}\left(\dfrac{y^{2}}{2}+\dfrac{1}{2y^{2}}\right)dy=\left[\dfrac{y^{3}}{6}-\dfrac{1}{2y}\right]_{1}^{2}$.

Evaluate

$\left(\dfrac86-\dfrac14\right)-\left(\dfrac16-\dfrac12\right)=\dfrac{17}{12}$.

EX 4

When it is not elementary

set up and evaluate numerically

Set up the length of $y=x^{2}$ from $x=0$ to $x=2$.

Differentiate

$\dfrac{dy}{dx}=2x$.

Build the integrand

$\sqrt{1+4x^{2}}$.

Write the integral

$L=\displaystyle\int_{0}^{2}\sqrt{1+4x^{2}}\,dx$.

Recognise the situation

This needs trigonometric substitution, which is outside the course. The set-up is the answer on a free-response question.

Evaluate numerically

$L\approx4.647$. A rough check: the straight chord from $(0,0)$ to $(2,4)$ has length $\sqrt{20}\approx4.47$, and the curve must be slightly longer. ✓

Watch out

The $1$ is inside the square root

$\sqrt{1+\left(f'\right)^{2}}$, not $1+\sqrt{\left(f'\right)^{2}}$. The second expression simplifies to $1+\lvert f'\rvert$ and is not an arc length.

Square the derivative, not the function

The integrand involves $f'$ only. The function's own values never appear in the arc-length formula.

Most arc-length integrals are not elementary

If the square root refuses to simplify, that is expected. Write the integral correctly and evaluate it numerically.

Check smoothness before integrating

A corner or a vertical tangent invalidates the $dx$ form. Switching to the $dy$ form often rescues the vertical-tangent case; a genuine corner requires splitting the curve.

Course map. Topic 8.13 · LO CHA-6.A — determine the length of a curve in the plane defined by a function, using a definite integral. BC-only.